Question

Two moles of a monoatomic ideal gas at 10 atm and 300 K is expanded isothermally and reversibly to a pressure of 2 atm. The absolute value of work done by the system is (in kJ) .......... (rounded off to two decimal places).

Hint:

For isothermal reversible expansion, the work depends only on the pressure ratio, not on absolute values. Always use natural logarithm $\ln(P_i/P_f)$.

Answer 1

Correct Answer: 7.9

Step 1: Recall isothermal work formula.
For a reversible isothermal process: \[ W = nRT \ln \left(\frac{V_f}{V_i}\right) = nRT \ln \left(\frac{P_i}{P_f}\right) \]
Step 2: Substitute values.
\[ n = 2 \, mol, R = 8.31 \, J/molK, T = 300 \, K \] \[ P_i = 10 \, atm, P_f = 2 \, atm \] \[ W = 2 \times 8.31 \times 300 \times \ln \left(\frac{10}{2}\right) \]
Step 3: Simplify terms.
\[ 2 \times 8.31 \times 300 = 4986 \] \[ \ln(5) \approx 1.609 \] \[ W = 4986 \times 1.609 = 8020.6 \, J \]
Step 4: Convert to kJ.
\[ W = 8.02 \, kJ \] Final Answer: \[ \boxed{8.02 \, kJ} \]